A solid hemisphere is mounted on a solid cyli | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $r$ be the common radius and $h$ be the height of the cylinder.The total surface area $S$ of the solid is the sum of the curved surface area o

Vedclass · Accepted Answer

$1: 1$

Vedclass · Answer

(A) Let $r$ be the common radius and $h$ be the height of the cylinder.The total surface area $S$ of the solid is the sum of the curved surface area of the hemisphere $(2\pi r^2)$,the curved surface area of the cylinder $(2\pi rh)$,and the base area of the cylinder $(\pi r^2)$.$S = 2\pi r^2 + 2\pi rh + \pi r^2 = 3\pi r^2 + 2\pi rh$From this,we can express $h$ in terms of $S$ and $r$:$h = \frac{S - 3\pi r^2}{2\pi r}$The volume $V$ of the solid is the sum of the volume of the hemisphere $(\frac{2}{3}\pi r^3)$ and the volume of the cylinder $(\pi r^2 h)$:$V = \pi r^2 h + \frac{2}{3}\pi r^3$Substituting the expression for $h$:$V = \pi r^2 \left( \frac{S - 3\pi r^2}{2\pi r} \right) + \frac{2}{3}\pi r^3$$V = \frac{1}{2} (Sr - 3\pi r^3) + \frac{2}{3}\pi r^3 = \frac{1}{2} Sr - \frac{3}{2}\pi r^3 + \frac{2}{3}\pi r^3 = \frac{1}{2} Sr - \frac{5}{6}\pi r^3$To find the maximum volume,differentiate $V$ with respect to $r$ and set it to $0$:$\frac{dV}{dr} = \frac{S}{2} - \frac{5}{2}\pi r^2 = 0$$S = 5\pi r^2$Now,substitute $S = 5\pi r^2$ back into the expression for $h$:$h = \frac{5\pi r^2 - 3\pi r^2}{2\pi r} = \frac{2\pi r^2}{2\pi r} = r$Thus,the ratio of the height of the cylinder to the common radius is $h:r = 1:1$.

$A$ solid hemisphere is mounted on a solid cylinder,both having equal radii. If the whole solid is to have a fixed surface area and the maximum possible volume,then the ratio of the height of the cylinder to the common radius is

Similar Questions

The greatest value of the real-valued function $f(x)=(x+1)^{1/3}-(x-1)^{1/3}$ on the interval $[0, 1]$ is:

Let $f(x) = \begin{cases} 3 - x, & 0 \le x < 1 \\ x^2 + \log_e b, & x \ge 1 \end{cases}$. The set of values of $b$ such that $f(x)$ has a local minimum at $x = 1$ is

If the function $f(x) = x^3 - 3(a - 2)x^2 + 3ax + 7$,for some $a \in R$,is increasing in $(0, 1]$ and decreasing in $[1, 5)$,then a root of the equation $\frac{f(x) - 14}{(x - 1)^2} = 0$ $(x \neq 1)$ is

The maximum value of $xy$ subject to $x+y=7$ is

Among all sectors of a fixed perimeter,choose the one with maximum area. Then,the angle at the centre of this sector (i.e.,the angle between the bounding radii) is

Vedclass Test Series

Exam Paper Generator

Online Exam Module